anonymous
  • anonymous
can't understand why integral of xe^((x^2)/2) is e^((x^2)/2) can someone explain how to calculate this integral please?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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UsukiDoll
  • UsukiDoll
\[\LARGE xe^{\frac{x^2}{2}}\] this one?
UsukiDoll
  • UsukiDoll
do you know u-substitution?
UsukiDoll
  • UsukiDoll
The u-substitution is an integration method that essentially involves using the chain rule in reverse.

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UsukiDoll
  • UsukiDoll
hmm let's try rewriting that fraction \[\LARGE xe^{\frac{1}{2}x^2}\] so we can see the exponent clearly.
UsukiDoll
  • UsukiDoll
we need to pick a u so that when we take that derivative of u we can substitute.. so what will be our u? Well we do see an x in front of the e.. Maybe we can have \[\LARGE u=\frac{1}{2}x^2 \]
UsukiDoll
  • UsukiDoll
so what is the derivative of \[\large \frac{1}{2}x^2 ? \]
UsukiDoll
  • UsukiDoll
we can use product rule to make this a bit easier. Let f(x) = 1/2 and g(x) = x^2 recall that the product rule is f(x)g'(x) +f'(x)g(x)
UsukiDoll
  • UsukiDoll
so what is the derivative of f(x) = 1/2 ?
UsukiDoll
  • UsukiDoll
similarly what is the derivative of \[\large g(x) = x^2 \]
anonymous
  • anonymous
so if \[u= (1/2)x^2 -> u' = x\]
UsukiDoll
  • UsukiDoll
yeah but we have to write this differently it will be du = x dx
anonymous
  • anonymous
so now we have \[xe^u dx\]
UsukiDoll
  • UsukiDoll
hmm how do I explain this.... we just have \[e^u \] and substitute the u back .. don't forget to add +C
UsukiDoll
  • UsukiDoll
remember du = x dx ? since there's nothing on the left hand side.... oh man .. I'm having trouble explaining this part. . . usually when there's nothing ... it's just blank like 1 sorry I'm in panic mode I can't figure out how to type this well
UsukiDoll
  • UsukiDoll
but we just have \[\LARGE e^u +C \] now substitute \[\LARGE u=\frac{1}{2}x^2 \] back into \[\LARGE e^u +C \]
anonymous
  • anonymous
ok, so \[xe^u dx -> xe^u (1/x) dy -> e^u du\] (+c...)
anonymous
  • anonymous
du
UsukiDoll
  • UsukiDoll
you don't have to write dx you don't have to write du
UsukiDoll
  • UsukiDoll
du = x dx it's just .... @ganeshie8 can you help me explain this? I know what's going on just can't put it into words
anonymous
  • anonymous
Ok, I think I understand, thank you very much!
UsukiDoll
  • UsukiDoll
but in the end you should have \[\LARGE e^u +C \] now plug in u back into this equation
UsukiDoll
  • UsukiDoll
and once we have our antiderivative, we can simply check by taking the derivative.
UsukiDoll
  • UsukiDoll
you should have something like \[\LARGE e^{\frac{1}{2}x^2} +C\]
UsukiDoll
  • UsukiDoll
now take the derivative of that equation...
UsukiDoll
  • UsukiDoll
that's how to check :)
anonymous
  • anonymous
so we get back to \[e^((x^2)/2) \]
anonymous
  • anonymous
yes, +c
UsukiDoll
  • UsukiDoll
mhm
UsukiDoll
  • UsukiDoll
if you take the derivative of your antiderivative, we should be getting that x in front of that e
anonymous
  • anonymous
Yes, I wrote the integral
UsukiDoll
  • UsukiDoll
\[\LARGE f(x) = e^{\frac{1}{2}x^2} +C \] \[\LARGE f'(x) = xe^{\frac{1}{2}x^2} +C \] because the derivative is \[\frac{1}{2}x^2 \rightarrow \frac{1}{2} (2x) + 0(x^2) \rightarrow x\] and that's placed in front of the e.
UsukiDoll
  • UsukiDoll
the derivative of any e^x is always in the front and the exponent part of e^x just stays there :)
anonymous
  • anonymous
Excellent. Thanks a lot!

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